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It's easy to see that the extremum in $(0,1)$ has the same magnitude as the one in $(k-1,k-2)$ and is more extreme than any of the other extrema. The extremum in $(0,1)$ occurs at $x_0=(1+o(1))/\ln k …

(CORRECTED EDITION) By mucking around with expansions like Igor suggested, I found $$ j \approx J(n,k)= en^{1/2} - \tfrac14\ln(n) -\tfrac12\ln(2\pi k)-\tfrac14 e^2-\tfrac12. $$ It seems good when …

This is not an answer but just some hints. Define $A(s)$ to be the quantity on the left and $B(s)$ to be the quantity on the right. Define $A(0)=B(0)=1$. Then it suffices to prove that $$ \sum_{s\g …

There is a chapter in Knuth and Greene, Mathematics for the analysis of algorithms, that explains how to estimate this type of thing. If $B$ is fixed and $n\to\infty$, the central limit theorem might …

Here's a very clunky approach that might not be close to a general solution. However, it is a proof for $r\le 99$. Define $p_r(n)$ by $$ \sum_{k=0}^n (-1)^k \binom {2n}k (n-k)^r = (-1)^n\binom{2n}{ …

What you have is the $n$-th difference operator applied to the sequence $(a_i)$. In particular, the value is 0 if $a_i=f(i)$ for any polynomial $f$ of degree less than $n$. The converse is also true …

Here's one from an old paper of mine. It has the property of being precise all the way from the middle to the end. Define $$ Y(x) = e^{x^2/2}\int_x^\infty e^{-t^2/2}dt. $$ Define $x=(2k-n)/\sqrt{n}$. …

For smallish $k$, we have $$ \binom{n}{n/2+k} \approx \binom{n}{n/2} \exp(-2k^2/n). $$ So $$\sum\sqrt{\binom{n}{n/2+k}} \approx \sqrt{\binom{n}{n/2}} \int_{-\infty}^\infty e^{-k^2/n}\,dk \approx …

It isn't true. Choose $n,a,k$ such that $n+a-k>0$, $n+a-2k<0$ and $k>1$. The sum has only one nonzero term which is not equal to the right side.

Following on from Carlo's answer, someone noted in response to one of my questions once that for a unimodal nonnegative function $f$, we have $$ \biggl| \sum_{i=0}^k f(i) - \int_0^k f(x)\,dx \biggr| …

Here is a confident guess, without a proof. The normal approximation of the binomial distribution gives the estimate $$ f(n,p) = \frac{\sqrt{2pq}}{\sqrt{\pi n}}.$$ Now, experimentally, for any fixed …

Expanding on the previous answers. I'm taking $\lambda$ and $\alpha$ to be constants which do not vary as $n\to\infty$. If $α<λ/(λ+1)$ then the sum is within a constant of the last term. In fact the …

If the value of the sum is $\frac13-\varDelta(k)$, then it appears that $\varDelta(k)$ satisfies the recurrence $$ (8k+4)\varDelta(k) = (7k-5)\varDelta(k-1) + k\varDelta(k-2).$$ Note that I didn't pro …

(Partial results.) For the case of integer ratio, there are only two sequences of 4 binomials in geometric progression for which the largest is at most $10^{17}$. Namely, 55,165,495,1485 found by Will …

Let $F(x)$ be the right side. You need that $F(x)=0$ for [corrected] $x\in\{-n,-n+1,\ldots,n-1,n\}$. Assume we have one of those $x$s. The summation can stop at $m=n-x$ as later terms are zero. For …